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Set up the equation for the arithmetic mean.

Mathematics

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2 parts
a

1.85 m\textbf{1.85 m}

b(ii)

1\textbf{1}

Step-by-step solution

5 steps

Here is the solution to question 1:

Part a· 2 steps

To calculate the value of xx, we use the formula for the arithmetic mean.

  1. Set up the equation for the arithmetic mean.

    The arithmetic mean is the sum of the heights divided by the number of students.

    1.50+x+1.453=1.60\frac{1.50 + x + 1.45}{3} = 1.60
  2. Solve for xx.

    Multiply both sides by 3:

    1.50+x+1.45=1.60×31.50 + x + 1.45 = 1.60 \times 3 2.95+x=4.802.95 + x = 4.80

    Subtract 2.95 from both sides:

    x=4.80−2.95x = 4.80 - 2.95 x=1.85x = 1.85

    The value of xx is 1.85 m\boxed{\textbf{1.85 m}}.

Part b(i)

Three main ways of describing sets are

  • Roster method (or Listing method): Listing all the elements of the set within curly braces, e.g., {1,2,3}\{1, 2, 3\}.
  • Set-builder notation (or Rule method): Describing the properties that the elements of the set must satisfy, e.g., {x∣x is an even number}\{x \mid x \text{ is an even number}\}.
  • Verbal description (or Word method): Describing the set in words, e.g., "the set of natural numbers less than 5".

Part b(ii)

1\textbf{1}

To find n(E∩P)n(E \cap P), we need to identify the elements that are common to both the set of even numbers (EE) and the set of prime numbers (PP).

  1. Identify the sets EE and PP.

    E={…,−4,−2,0,2,4,6,… }E = \{\dots, -4, -2, 0, 2, 4, 6, \dots\} (all integers divisible by 2)
    P={2,3,5,7,11,… }P = \{2, 3, 5, 7, 11, \dots\} (natural numbers greater than 1 with no positive divisors other than 1 and itself)

  2. Find the intersection E∩PE \cap P.

    The only number that is both even and prime is 2.
    So, E∩P={2}E \cap P = \{2\}.

  3. Determine the number of elements in the intersection.

    n(E∩P)n(E \cap P) is the number of elements in the set {2}\{2\}, which is 1.
    n(E∩P)=1n(E \cap P) = \boxed{\textbf{1}}.

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